(3m+1)(m=5)

Solution for (3m+1)(m=5) equation:

(3m+1)(m=5)

We move all terms to the left:

(3m+1)(m-(5))=0

We multiply parentheses ..

(+3m^2-15m+m-5)=0

We get rid of parentheses

3m^2-15m+m-5=0

We add all the numbers together, and all the variables

3m^2-14m-5=0

a = 3; b = -14; c = -5;
Δ = b2-4ac
Δ = -142-4·3·(-5)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-16}{2*3}=\frac{-2}{6}
=-1/3
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+16}{2*3}=\frac{30}{6}
=5
$